Answer:
i think its b
Explanation:
i did the test and got it right
Answer:
-26
Explanation:
The given binary number is 1110 0101. Also given that the signed binary number is represented using one's compliment.
We begin by computing the 1s complement representation of 1110 0101 by inverting the bits: 00011010
Converting 00011010 to decimal, it corresponds to 26.
So the 1s complement of the original number is 26. This means that the original number was -26.
Answer:
C++ code is given below
Explanation:
#include<iostream>
#include <cstring>
using namespace std;
int housekeeping(string EOFNAME);
int mainLoop(string name,string EOFNAME);
int finish();
void main()
{
string name;
string EOFNAME = "ZZZZ";
cout << "enter the name" << endl;
cin >> name;
if (name != EOFNAME)
{
housekeeping(EOFNAME);
}
if (name != EOFNAME)
{
mainLoop(name , EOFNAME);
}
if (name != EOFNAME)
{
finish();
}
system("pause");
}
int housekeeping(string EOFNAME)
{
cout << "enter first name " << EOFNAME << " to quit " << endl;
return 0;
}
int mainLoop(string name, string EOFNAME)
{
int hours;
int rate,gross;
int DEDUCTION = 45;
int net;
cout << "enter hours worked for " << name << endl;
cin >> hours;
cout << "enter hourly rate for " << name << endl;
cin >> rate;
gross = hours*rate;
net = gross - DEDUCTION;
if (net > 0)
{
cout << "net pay for " << name << " is " << net << endl;
}
else
{
cout << "dedections not covered.net is 0.";
}
cout << "enter next name or " << EOFNAME << " to quit" << endl;
cin >> name;
return 0;
}
int finish()
{
cout << "end of job"<<endl;
return 0;
}
Answer:
#include<iostream>
using namespace std;
void main()
{
int townA_pop,townB_pop,count_years=1;
double rateA,rateB;
cout<<"please enter the population of town A"<<endl;
cin>>townA_pop;
cout<<"please enter the population of town B"<<endl;
cin>>townB_pop;
cout<<"please enter the grothw rate of town A"<<endl;
cin>>rateA;
cout<<"please enter the grothw rate of town B"<<endl;
cin>>rateB;
while(townA_pop < townB_pop)//IF town A pop is equal or greater than town B it will break
{
townA_pop = townA_pop +( townA_pop * (rateA /100) );
townB_pop = townB_pop +( townB_pop * (rateB /100) );
count_years++;
}
cout<<"after "<<count_years<<" of years the pop of town A will be graeter than or equal To the pop of town B"<<endl;
}
Explanation:
c. would be the best answer.
